, 2020

Solution:

Let $X, Y$ be on $AB, AC$ such that $CX \parallel BE$ and $BY \parallel CF$. Then $BX = BC = CY$. Let $Z$ be the midpoint of $XY$. Then $\overrightarrow{MZ} = \frac{1}{2}(\overrightarrow{BX} + \overrightarrow{CY})$, which bisects the angle between $BX$ and $CY$ because they have the same length. Therefore $MZ \parallel AD$. Furthermore, by similar triangles we have
$$AE \cdot AX = AB \cdot AC = AF \cdot AY$$
This rearranges to $\frac{AE}{AF} = \frac{AY}{AX}$, so $EF \parallel XY$. Therefore $Z$ is the intersection of the lines in the problem statement. Then
$$\frac{\sin \angle BZX}{\sin \angle CZY} = \frac{BX \frac{\sin \angle ZBX}{XZ}}{CY \frac{\sin \angle ZCY}{YZ}} = 1$$
iff $Z \in (ABC)$, so $XY$ is the external angle bisector of $\angle BZC$ iff $Z \in (ABC)$. Thus if $P = AD \cap XY$, $P \in (ABC)$ if $Z \in (ABC)$. Additionally the spiral similarity from $BX$ to $CY$ gives $L_A Z \perp XY$ where $L_A$ is the midpoint of $\operatorname{arc} BAC$, so if $P \in (ABC)$ then $Z$ must be on $(ABC)$ because $\angle L_A ZP = 90^\circ$. Therefore $Z \in (ABC)$ iff $P \in (ABC)$.
From the previous length computation, we know that an inversion at $A$ with radius $\sqrt{AB \cdot AC}$ composed with reflection about $AD$ will send $X$ and $Y$ to $E$ and $F$. We have $P \in (ABC)$ iff its image under the inversion is $D$, but since $P$ was defined as $AD \cap XY$ this is true iff $(AEDF)$ is cyclic. Since $ABC$ is scalene and $AE \neq AF$, this is true iff $DE = DF$.

Solution:

Let $L_A$ be the midpoint of arc $BAC$ and let $M_A$ be diametrically opposite $L_A$. Let $EF$, $AL_A$, and $BC$ meet at $T$ so $\angle DAT = 90^\circ$; note that $DE = DF$ iff $DN \perp EF$, which is equivalent to $(TAND)$ being cyclic. Let $AN \cap (ABC) = X$ and $XM \cap (ABC) = Y$, and let $Y'$ be the reflection of $Y$ over $L_A M_A$ with similarly $X'$ the reflection of $X$ over $L_A M_A$. We wish to show $N \in (TAND)$ iff $XY \parallel AM_A$.
We claim $AY' \parallel EF$. By projecting $-1 = (B, C; M, \infty) \stackrel{X}{=} (B, C; Y, X')$ and reflecting over $L_A M_A$, we find $(X, Y'; B, C) = -1$. Then projecting through $A$ gives $(N, AY' \cap EF; F, E) = -1$, and since $N$ is the midpoint of $EF$ we find $AY' \parallel EF$.
Now $(TAND)$ cyclic iff $\measuredangle DAN = \measuredangle DTN$, and $\measuredangle DTN = \measuredangle YY'A$ by the parallel lines. But we have $\measuredangle DAN = \measuredangle M_A AX$, so $\operatorname{arcs} M_A X$ and $YA$ are equal iff $(TAND)$ is cyclic. Thus $XY \parallel AM_A$ iff $DE = DF$ as desired.